Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{4t + 36}{-7t - 42} \div \dfrac{3t + 27}{t^2 + 10t + 24} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{4t + 36}{-7t - 42} \times \dfrac{t^2 + 10t + 24}{3t + 27} $ First factor the quadratic. $r = \dfrac{4t + 36}{-7t - 42} \times \dfrac{(t + 6)(t + 4)}{3t + 27} $ Then factor out any other terms. $r = \dfrac{4(t + 9)}{-7(t + 6)} \times \dfrac{(t + 6)(t + 4)}{3(t + 9)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ 4(t + 9) \times (t + 6)(t + 4) } { -7(t + 6) \times 3(t + 9) } $ $r = \dfrac{ 4(t + 9)(t + 6)(t + 4)}{ -21(t + 6)(t + 9)} $ Notice that $(t + 9)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ 4\cancel{(t + 9)}(t + 6)(t + 4)}{ -21\cancel{(t + 6)}(t + 9)} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $r = \dfrac{ 4\cancel{(t + 9)}\cancel{(t + 6)}(t + 4)}{ -21\cancel{(t + 6)}\cancel{(t + 9)}} $ We are dividing by $t + 9$ , so $t + 9 \neq 0$ Therefore, $t \neq -9$ $r = \dfrac{4(t + 4)}{-21} $ $r = \dfrac{-4(t + 4)}{21} ; \space t \neq -6 ; \space t \neq -9 $